Theory of relativity - proton entering a medium

[Saitama]

Homework Statement

(see attachment)
Given data:
1.The speed of light in empty space is $3 \times 10^8 m/s$.
2.The mass of the proton is $1.67 \times 10^{-27} kg$ .
3.The speed of light in water is $2.3 \times 10^8 m/s$.
4.When photon emission begins to occur the initial proton, the final proton, and the emitted photon all travel in the same direction.

Homework Equations

The Attempt at a Solution

I don't have any knowledge about theory of relativity but I think this question can be done using the info given in the question.

Initial energy of proton, $E_i=\sqrt{c^2p^2+m_0^2c^4}$
Final energy of proton, $E_f=\sqrt{c^2p_1^2+m_0^2c^4}$
where $m_0$ is the rest mass of proton, $p$ is the initial momentum of proton and $p_1$ is the final momentum when it enters the water.
Using conservation of energy,
$$E_i=E_{photon}+E_f$$
Using conservation of linear momentum,
$$p=p_1+p_2$$
where $p_2$ is the momentum of emitted photon.
I can find $p_2$ in terms of $p$ and $p_1$ but I am not sure what I am supposed to do here. I can substitute p_2 in the energy equation but what am I supposed to do with that equation?

Any help is appreciated. Thanks!

 

[Saitama]

Anyone?

 

[haruspex]

You've not used the equation E_ph = c_m|p₂|.
Try approximating Ei, Ef on the basis that p << m₀c.

 

[Saitama]

Thanks Haruspex for your reply!
I used the equation E=c_m|p|, took the approximation and substituted them in the energy equation. I ended up with a relation between p and p₁.
$$p=2m_o^2c^2c_m-p_1$$
What am I supposed to do next?

 

[haruspex]

I used the equation E=c_m|p|, took the approximation and substituted them in the energy equation. I ended up with a relation between p and p₁.
$$p=2m_o^2c^2c_m-p_1$$

That can't be right - it's dimensionally inconsistent. (I get $p=2m_oc_m-p_1$.)
How small can p₁ be?

 

[Saitama]

That can't be right - it's dimensionally inconsistent. (I get $p=2m_oc_m-p_1$.)
How small can p₁ be?

Woops, I forgot to cancel $m_oc^2$ while taking the approximation.

smallest value of p1 can be zero?

 

[haruspex]

smallest value of p1 can be zero?

That's what I would assume.

 

[Saitama]

ok so I plugged in the values. I get $7.682 \times 10^{-19}$ but this is wrong. :-(

 

[haruspex]

Having ended up setting p1 to zero, I realize that we didn't need to do the approximation. Try putting p1=0 but not approximating anything. I think you should get 1.86 10^-18. Is that better?

 

[Saitama]

Having ended up setting p1 to zero, I realize that we didn't need to do the approximation. Try putting p1=0 but not approximating anything. I think you should get 1.86 10^-18. Is that better?

It still says that its incorrect and now I can't try submitting an answer because all my tries are over. :-(

 

[haruspex]

It still says that its incorrect and now I can't try submitting an answer because all my tries are over. :-(

And I'm out of ideas - sorry.

 

[Saitama]

And I'm out of ideas - sorry.

The answers were published today. The answer given is $3.58 \times 10^{-37}$.

 

[haruspex]

The answers were published today. The answer given is $3.58 \times 10^{-37}$.

Hmm.. that's numerically about the square of what we calculated. Any idea how that result is obtained?
EDIT: Just ran a check I wish I'd tried before: with my result, the implied initial speed of the proton > c. I guess setting p1=0 is not the way.

 

[Saitama]

I guess setting p1=0 is not the way.

Then how?

 

[sankalpmittal]

Then how?

My knowledge regarding general relativity is also meagre but the question asks the minimum magnitude of proton's momentum and not the maximum. You should rather choose for maximum value of p₁. Its certainly not 0. What is it ?

Not quite sure of that. You can not be.

Edit: Aside, I really suspect that approximation will work.

Ok,ok. A more clear approach will be to eliminate one of the variables(not p) from momentum conservation , and the plug in energy conservation with two unknowns. Just isolate p in terms of p1 or p2 and analyze what value you can set for p1 or p2 for minimum p. What do you get ? Please do not approximate anything!

Really, you should get quadratic sort of equation.

 

[haruspex]

My knowledge regarding general relativity is also meagre but the question asks the minimum magnitude of proton's momentum and not the maximum. You should rather choose for maximum value of p₁. Its certainly not 0. What is it ?

p1 is the proton's momentum after the event. The approximation we tried to begin with suggested that the larger that was, the larger the initial momentum needed to be, so it seemed like the smallest p would go with p1=0. Having selected that, it was easy to rerun the calculation without approximating.

Ok,ok. A more clear approach will be to eliminate one of the variables(not p) from momentum conservation , and the plug in energy conservation with two unknowns. Just isolate p in terms of p1 or p2 and analyze what value you can set for p1 or p2 for minimum p. What do you get ? Please do not approximate anything!

Really, you should get quadratic sort of equation.

It gives a quartic, which is why I went for approximation. There might be some cancellation (of p-p1), but the algebra gets a bit heavy.

 

[sankalpmittal]

It gives a quartic, which is why I went for approximation. There might be some cancellation (of p-p1), but the algebra gets a bit heavy.

A quartic ? That will be bit messy. Ok, but you know that approximation was not giving the correct answer, right?

Can you state that quartic, in terms of p and p₁ here? My exams are over on this very day and I am feeling too exhausted to do this. Or rather, we should wait for the OP to post...

 

[haruspex]

A quartic ? That will be bit messy. Ok, but you know that approximation was not giving the correct answer, right?

It wasn't just wrong, it was out by about 18 orders of magnitude.

Can you state that quartic, in terms of p and p₁ here? My exams are over on this very day and I am feeling too exhausted to do this. Or rather, we should wait for the OP to post...

Initial energy of proton, $E_i=\sqrt{c^2p^2+m_0^2c^4}$
Final energy of proton, $E_f=\sqrt{c^2p_1^2+m_0^2c^4}$
where $m_0$ is the rest mass of proton, $p$ is the initial momentum of proton and $p_1$ is the final momentum when it enters the water.
Using conservation of energy,
$$E_i=E_{ph} +E_f$$
Using conservation of linear momentum, with p₂ as the momentum of the photon
$$p=p_1+p_2$$
$E_{ph} = c_m|p_2|$
Putting it all together:
$\sqrt{c^2p^2+m_0^2c^4} - \sqrt{c^2p_1^2+m_0^2c^4} = c_m(p - p_1)$
Square both sides, isolate the surd on one side, square again. Actually it's not a quartic, but rather worse. OTOH, there may well be a factor of (p-p₁) to take out.

 

[sankalpmittal]

It wasn't just wrong, it was out by about 18 orders of magnitude.

Initial energy of proton, $E_i=\sqrt{c^2p^2+m_0^2c^4}$
Final energy of proton, $E_f=\sqrt{c^2p_1^2+m_0^2c^4}$
where $m_0$ is the rest mass of proton, $p$ is the initial momentum of proton and $p_1$ is the final momentum when it enters the water.
Using conservation of energy,
$$E_i=E_{ph} +E_f$$
Using conservation of linear momentum, with p₂ as the momentum of the photon
$$p=p_1+p_2$$
$E_{ph} = c_m|p_2|$
Putting it all together:
$\sqrt{c^2p^2+m_0^2c^4} - \sqrt{c^2p_1^2+m_0^2c^4} = c_m(p - p_1)$
Square both sides, isolate the surd on one side, square again. Actually it's not a quartic, but rather worse. OTOH, there may well be a factor of (p-p₁) to take out.

Yes ! Thanks Haruspex !

You are correct. I solved it further and the factor (p-p₁) canceled out with the calculation. I have eliminated the radicals and have obtained the following equation :

c⁴(p+p₁)² + c_m⁴(p-p₁)² - 2c²c_m²p²= 2c_m²c²p₁² + 4c_m²m_o²c⁴

I guess that we will have mess with this equation to find minimum magnitude of p. Since all terms are basically squared, setting p₁=0 , in this should do (but it did not ?).

There must be usage of some mathematical principle to find the magnitude of p1 so as to minimize p.

No, I think setting p1=0 should do. I get a quadratic equation in p.
I am sorry. There must be another way.

Nonetheless, no doubt this is the only way.

 

[Saitama]

Yes ! Thanks Haruspex !

You are correct. I solved it further and the factor (p-p₁) canceled out with the calculation. I have eliminated the radicals and have obtained the following equation :

c⁴(p+p₁)² + c_m⁴(p-p₁)² - 2c²c_m²p²= 2c_m²c²p₁² + 4c_m²m_o²c⁴

I guess that we will have mess with this equation to find minimum magnitude of p. Since all terms are basically squared, setting p₁=0 , in this should do (but it did not ?).

There must be usage of some mathematical principle to find the magnitude of p1 so as to minimize p.

No, I think setting p1=0 should do. I get a quadratic equation in p.

Nonetheless, no doubt this is the only way.

Setting p1=0 doesn't give the right answer. Oh my, how did you get the patience to do that dirty algebra. :tongue:

 

[sankalpmittal]

Setting p1=0 doesn't give the right answer. Oh my, how did you get the patience to do that dirty algebra. :tongue:

Then rather, open up the equation which I got in my previous post, and write it in the form : ap²+bp+c=0, where c contains all the terms of p₁. The the discriminant b²-4ac≥0, should give you some sort of range of values of p₁. And that was not very much algebra. Term p-p₁ canceled out as marked by haruspex.

 

[Saitama]

Then rather, open up the equation which I got in my previous post, and write it in the form : ap²+bp+c=0, where c contains all the terms of p₁. The the discriminant b²-4ac≥0, should give you some sort of range of values of p₁. And that was not very much algebra. Term p-p₁ canceled out as marked by haruspex.

I tried to differentiate the energy equation (consisting of surds) wrt $p_1$ to find the minimum value of p. I ended up with a quadratic in $p_1$. Found two values of ##p_1# but none of them gave the right answer when I substituted them in the energy equation.

 

[sankalpmittal]

I tried to differentiate the energy equation (consisting of surds) wrt $p_1$ to find the minimum value of p. I ended up with a quadratic in $p_1$. Found two values of ##p_1# but none of them gave the right answer when I substituted them in the energy equation.

Can you show the working? :tongue:

And please try the discriminant method which I cited in my previous post. I think it will give you the answer. Setting up a maxima or minima is sometimes problematic. Did you analyze d²p/dp₁² also ?

Also if this does not work follow:

ap²+bp+c=0

Now use the quadratic formula and isolate p in terms of p₁. One root of the two corresponding roots will be minimum value of p. Then from that root you can make p1 self adjust so that p will be surely minimum.

If all those method also not work (except that you do a careless mistake), then you will have to contact other HHs of PF. This is getting really messy now.

 

[TSny]

I'm not understanding something here. So, please help me out.

What if you go to the reference frame in which the proton is initially at rest. So, initially, the total energy in this frame is just the rest-mass energy of the proton. How can the proton in this frame emit a photon and conserve energy?

The only way that I can imagine the proton emitting a photon is if the proton interacts with the background water. But then you would need to include terms for the momentum and energy transferred to the water.

 

[voko]

I'm not understanding something here. So, please help me out.

What if you go to the reference frame in which the proton is initially at rest. So, initially, the total energy in this frame is just the rest-mass energy of the proton. How can the proton in this frame emit a photon and conserve energy?

The only way that I can imagine the proton emitting a photon is if the proton interacts with the background water. But then you would need to include terms for the momentum and energy transferred to the water.

Not only that, the very concept of "momentum" is controversial for photons moving in a medium. There are two conflicting definitions and over 100 years of debate which one is correct.

 

[haruspex]

I'm not understanding something here. So, please help me out.

What if you go to the reference frame in which the proton is initially at rest. So, initially, the total energy in this frame is just the rest-mass energy of the proton. How can the proton in this frame emit a photon and conserve energy?

In that frame, the proton sitting minding its own business got whacked by a tsunami. No shortage of energy there.

 

[haruspex]

Cracked it! It also occurred to me in the middle of the night that I should just differentiate the surd equation wrt p₁ and set dp/dp₁=0. That gives p = m₀cc_m/√(c²-c_m²) = p₁. At first sight that looks useless, since it gives p = p₁, but then it dawned on me that the likely alternative to setting p₁=0 is to set p-p₁ very small. Indeed, that corresponds to the photon getting a vanishingly small amount of energy. (With that assumption in the first place, we could have expanded the surds on the basis of small δp = p-p₁ and got the result quite quickly.)
Numerically it gives 5.98.10^-19. Why so different from the given answer? Because the given answer is wrong, 3.58.10^-37 being the numerical value of p².

 

[TSny]

In that frame, the proton sitting minding its own business got whacked by a tsunami. No shortage of energy there.

OK, so the proton must interact with the water, and the "tsunami" transfers energy to the proton/photon system? How was that taken into account?

 

[voko]

For the record, I have tried a constrained minimization approach, and that got me zero momentum for the photon.

 

[TSny]

Is the entire interaction of the proton with the water somehow taken into account by just using the dispersion relation E = c_mp for the photon?

 

[haruspex]

Is the entire interaction of the proton with the water somehow taken into account by just using the dispersion relation E = c_mp for the photon?

That's my understanding. I.e. it's not a 'solid' collision between the proton and water molecules, but the subtler relativistic interaction with a medium with index < 1. Not sure whether it's important that a proton carries charge - probably.

 

[sankalpmittal]

Cracked it! It also occurred to me in the middle of the night that I should just differentiate the surd equation wrt p₁ and set dp/dp₁=0. That gives p = m₀cc_m/√(c²-c_m²) = p₁. At first sight that looks useless, since it gives p = p₁, but then it dawned on me that the likely alternative to setting p₁=0 is to set p-p₁ very small. Indeed, that corresponds to the photon getting a vanishingly small amount of energy. (With that assumption in the first place, we could have expanded the surds on the basis of small δp = p-p₁ and got the result quite quickly.)
Numerically it gives 5.98.10^-19. Why so different from the given answer? Because the given answer is wrong, 3.58.10^-37 being the numerical value of p².

Same answer I get. But what I did was to write the equation in post #19 in the form, ap²+bp+c=0. Then there will be two corresponding roots of with I chose the one which seemed minimum, yet doubtful. So I considered both the roots. I differentiated both the roots w.r.t p₁, obviously putting dp/dp₁=0 and solved. Obviously I checked that if d²p/dp₁²>0 for minima. I got, I think values, out of one seems to be following the minima condition. That was the answer same as what you got. And now, the copy on which I did this all is lost.

 

[TSny]

The expression $\sqrt{c^2p^2+m_0^2c^4} - \sqrt{c^2p_1^2+m_0^2c^4} = c_m(p - p_1)$ can be rearranged as $\sqrt{c^2p^2+m_0^2c^4} - c_mp = \sqrt{c^2p_1^2+m_0^2c^4} -c_mp_1$

Measuring $p$ in units of $m_0c$ and measuring $c_m$ in units of $c$ we have

$\sqrt{p^2+1} - c_mp = \sqrt{p_1^2+1} - c_mp_1$

Thus, defining $f(p) = \sqrt{p^2+1} - c_mp$ we have $f(p) = f(p_1)$.

A plot of $f(p)$ for water ($c_m = .767$) is attached. The intersection of a horizontal line with the graph would correspond to possible initial and final momenta of the proton, with the initial momentum being the larger momentum. The minimum of the function occurs where the initial proton momentum would have its minimum value and still be able to emit a “soft” photon. This is seen to occur near $p = 1.195$ (or, $5.98 \times 10^{-19}$ kgm/s). This, of course, agrees with haruspex’s calculation.

In a vacuum, we would have c_m = 1. The attachment shows a plot of f(p) for this case and we can see the function no longer takes on a minimum. So, as expected, the proton cannot emit a photon and conserve energy-momentum in a vacuum.

In fact, it is easy to show that the value of p that minimizes f(p) corresponds to the proton having a speed of c_m. So, only when the proton enters the water with a speed greater than the speed of light in water can the proton emit a photon.

So I guess this has has something to do with Cherenkov radiation. But I wish I understood better why simply using $E_{photon} = c_mp_{photon}$ for the photon in water automatically takes into account transfer of energy and momentum with the water. I had never seen this before.

 

[Saitama]

Thanks Haruspex and TSny for your help. Your answers are correct. The solution for this problem was posted today. The solution reaches the same answer as you. And as TSny mentioned, the solution too gives a link to Cherenkov radiation. I will quote the solution:

First simplify by considering all the particles as moving in the x-direction. The initial particle energy and momentum is a point $(p_i,E_i)$ on the curve $E_{prot}=\sqrt{c^2 p^2+m_{prot}^2c^4}$. So is the final proton energy and momentum a point $(p_f,E_f)$, with $p_f<p_i$ and $E_f<E_i$. If energy and momentum can be conserved when emitting a photon, these two points must be able to be connected by a line with slope $c_m$. The lowest value of $p_i$ where this can happen is where the slope of the $E(p)$ curve for the proton equals $c_m$. Hence we have $\frac {dE} {dp}|_{p_{min}}=c_m$ as our condition for the lowest value $p_{min}$ of the momentum for the initial proton. Algebraically solving this equation for $p_{min}$ yields the answer of $5.98 \times 10^{-19}~kg~m/s$.Finally, the pretty blue light produced by high energy particles hitting water is called Cherenkov light and can be seen here.

I don't understand the part when it says that the points connect by a line with slope c_m. From where did this come from?

Measuring $p$ in units of $m_0c$ and measuring $c_m$ in units of $c$ we have
$$\sqrt{p^2+1} - c_mp = \sqrt{p_1^2+1} - c_mp_1$$

Can you please explain this to me?

 

[TSny]

Start with
$\sqrt{c^2p^2+m_0^2c^4} - \sqrt{c^2p_1^2+m_0^2c^4} = c_m(p - p_1)$

and divide through by $m_0c^2$. You will find that the equation simplifies to

$\sqrt{[p/(m_0c)]^2+1} - \sqrt{[p_1/(m_0c)]^2+1} = (c_m/c)[(p/(m_0c) - p_1/(m_0c)]$

So, if you measure p in units of $m_0c$ and measure $c_m$ in units of $c$, you get

$\sqrt{p^2+1} - \sqrt{p_1^2+1} = c_m(p - p_1)$

Going back to the original equation and letting $E(p) = \sqrt{c^2p^2+m_0^2c^4}$, you have $E(p) - E(p_1) = c_m(p-p_1)$. Or,
$$\frac{E(p) - E(p_1)}{p-p_1} = c_m$$
The left-hand side is the slope of the line that connects two points on the $E(p)$ vs $p$ graph corresponding to initial and final momenta ($p$ and $p_1$) of the proton. So the above equation says that the slope of the line must be equal to $c_m$ if energy and momentum are to be conserved. Thus if you draw any straight line of slope $c_m$ such that it passes through two points of the graph of $E(p)$, the two points will correspond to possible initial and final momenta of the proton such that energy and momentum will be conserved in photon emission. If you draw the line of slope $c_m$ such that it is tangent to the graph of $E(p)$, then you will have the case where the initial and final momenta of the proton are equal and the emitted photon would have zero energy. The slope of the tangent line is the derivative $E\:'(p)$. But $E\:'(p)$ can easily be shown to equal the speed of the proton. So, the minimum speed of the proton for photon emission is $c_m$.